Question

Lim x ---> -1 (x³ - 4x² - 5x) / (-2 + raiz(3-x))

Answer 1

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Calcular o limite:

$\mathsf{L=\underset{x\to -1}{\ell im}~\dfrac{x^3-4x^2-5x}{-2+\sqrt{3-x}}}$


Multiplique o numerador e o denominador pelo conjugado do denominador, que é  $\mathsf{-2-\sqrt{3-x}:}$

$\mathsf{L=\underset{x\to -1}{\ell im}~\dfrac{(x^3-4x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}{\big(\!\!-2+\sqrt{3-x}\,\big)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}}\\\\\\ \mathsf{L=\underset{x\to -1}{\ell im}~\dfrac{(x^3-4x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}{(-2)^2-\big(\sqrt{3-x}\,\big)^2}}\\\\\\ \mathsf{L=\underset{x\to -1}{\ell im}~\dfrac{(x^3-4x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}{4-(3-x)}}\\\\\\ \mathsf{L=\underset{x\to -1}{\ell im}~\dfrac{(x^3-4x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}{4-3+x}}\\\\\\ \mathsf{L=\underset{x\to -1}{\ell im}~\dfrac{(x^3-4x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}{x+1}}$



Temos uma indeterminação  "0/0". Isto significa que  x = – 1  é raiz do polinômio que aparece no numerador, e portanto, podemos fatorar o numerador por  x + 1. Observe:

x³ – 4x² – 5x

= x³ + x² – x² – 4x² – 5x

= x³ + x² – 5x² – 5x

= x² · (x + 1) – 5x · (x + 1)

= (x + 1) · (x² – 5x)


e o limite fica

$\mathsf{=\underset{x\to -1}{\ell im}~\dfrac{(x+1)\cdot (x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}{x+1}}$


Simplificando o fator comum  x + 1  que aparece no numerador e no denominador,

$\mathsf{=\underset{x\to -1}{\ell im}~(x^2-5x)\cdot \big(\!\!-2-\sqrt{3-x}\,\big)}\\\\\\ \mathsf{=\big((-1)^2-5\cdot (-1)\big)\cdot \big(\!\!-2-\sqrt{3-(-1)}\,\big)}\\\\ \mathsf{=(1+5)\cdot \big(\!\!-2-\sqrt{3+1}\,\big)}\\\\ \mathsf{=6\cdot \big(\!\!-2-\sqrt{4}\,\big)}\\\\ \mathsf{=6\cdot (-2-2)}\\\\ \mathsf{=6\cdot (-4)}$

$\mathsf{=-\,24}$          ✔


∴     $\boxed{\begin{array}{c}\mathsf{\underset{x\to -1}{\ell im}~\dfrac{x^3-4x^2-5x}{-2+\sqrt{3-x}}=-\,24} \end{array}}$          ✔


Bons estudos! :-)